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A solution prepared by dissolving a 2.50 g sample of an unknown compound dissolved in 34.0 g of benzene, $C_{6} H_{6}$ , boils 1.38° C higher than pure benzene. Which expression gives the molar mass of the unknown compound?
Compound $K_{b}$
$C_{6} H_{6}$ ${2.53}^{\circ} C . m^{- 1}$

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$2.50 \times 10^{3} \times \frac{2.53}{1.38} \times 2.53$

$1.38 \times \frac{34.0}{2.53} \times 2.50$

$2.53 \times \frac{2.50}{1.38}$

$2.50 \times 10^{3} \times \frac{2.53}{1.38} \times \frac{1}{34}$

answer is C.

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Detailed Solution

Boling temperature change,

$Δ T_{b} = K_{b} \times m$

$1.38 = 2.53 \times \frac{\frac{2.50}{M}}{34} \times 1000$

$\Rightarrow M = 2.50 \times 10^{3} \times \frac{2.53}{1.38} \times \frac{1}{34}$

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